# Proving partial order set in detail

Let $\le$ be a partially ordered set on the set $M$ and $f: M \to N$ bijective. Show that there's for $n, n' \in N$ a partially orderd set on $N$ defined:

$$n\le n'\iff f^{-1}(n)\le f^{-1}(n')$$

I need help on showing reflexivity, antisymmetry and transitivity in detail. This is what I can do:

(I) reflexivity, $\forall n\in N:(n,n)\in N$

$n\le n\iff f^{-1}(n)\le f^{-1}(n)$

This should be true, but how do I show that this is true and reflexive?

(II) antisymmetry, $\forall n,n'\in N:(n,n')\in R\implies(n',n)\in R$

$n \le n' \iff f^{-1}(n) \le f^{−1}(n') = n' ≤ n \iff f^{-1}(n') \le f^{−1}(n)$

and again this seems true, do I maybe have to do it differently?

(III) transitivity, $\forall n,n',n''\in N:(n,n')\in R \land (n',n'')\in R\implies(n,n'')\in R$

$n \le n' \iff f^{-1}(n) \le f^{−1}(n') \land n' \le n'' \iff f^{-1}(n') \le f^{−1}(n'')=n ≤ n' \le n'' \iff f^{-1}(n) \le f^{−1}(n') \le f^{−1}(n'')\implies n' \le n''' \iff f^{-1}(n') \le f^{−1}(n''')$

This was most of the work, but to me it still seems to be not enough.

If I'm missing something or doing this totally wrong then I'd be happy about an explainantion! If I'm right, but missing something to really show I,II and III then a solution to one of those could be enough for me! Thanks in advance.

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What is $f^{-}$? – William May 5 '13 at 12:50
f^(-1), I couldn't get the 1 up there, sorry! – nullmoon May 5 '13 at 12:59
That would be $f^{-1}$ (for future reference). We'll likewise use the braces for subscripts, numerators and denominators of fractions, and so forth. – Cameron Buie May 5 '13 at 13:04
Please check to see that I interpreted everything correctly. – Brian M. Scott May 5 '13 at 13:07
Looks right to me, thank you! – nullmoon May 5 '13 at 13:17

Mostly, this is just a matter of running through the details. This is made more difficult by the confusing notation, so I will distinguish between the original partial order on $M$ and our new one on $N$ by $\le_M$ and $\le_N$, respectively.
(I) Take $n\in N$. We necessarily have $f^{-1}(n)=f^{-1}(n)$, so since $\le_M$ is reflexive on $M$, then we have $f^{-1}(n)\le_Mf^{-1}(n),$ so $n\le_Nn$ by definition.
(II) Suppose that $n\le_Nn'$ and $n'\le_Nn$ for some $n,n'\in N$. By definition, $f^{-1}(n)\le_Mf^{-1}(n')$ and $f^{-1}(n')\le_Mf^{-1}(n)$, so since $\le_M$ is antisymmetric on $M$, then we have $f^{-1}(n)=f^{-1}(n')$. Since $f$ is a bijection, it follows that $n=n'$.
(III) Suppose that $n\le_Nn'$ and $n'\le_Nn''$ for some $n,n',n''\in N$. By definition, $f^{-1}(n)\le_Mf^{-1}(n')$ and $f^{-1}(n')\le_Mf^{-1}(n'')$, so since $\le_M$ is transitive on $M$, then we have $f^{-1}(n)\le_Mf^{-1}(n'')$, whence $n\le_Nn''$ by definition.
No problem! It's worth noting that if $g:N\to M$ is any injection, and $R$ is any relation on $M$, we may define a relation $S$ on $N$ by $x\:S\:y$ if and only if $g(x)\:R\:g(y)$. $S$ will have a given property of relations (e.g.: antisymmetry, comparability, trichotomy, irreflexivity) on $N$ if $R$ has the corresponding property of relations on $M$. If $g$ is a bijection, then $S$ will have a given property of relations on $N$ if and only if $R$ has the corresponding property of relations on $M$. – Cameron Buie May 5 '13 at 13:26